1. The problem asks to find the vertex, axis of symmetry, and y-intercept for given quadratic functions and verify your answers. 2. For #1, you wrote: - Vertex (V): (2, -1) - Axis of Symmetry (AOS): x = 2 - y-intercept: 1 3. For #2, you wrote: - Vertex (V): (-3, 2) - Axis of Symmetry (AOS): x = -3 - y-intercept: 2 4. For #3, the function is $y = 3x^2 + 2x$. - Axis of symmetry formula: $x = -\frac{b}{2a}$ - Here, $a=3$, $b=2$, so $x = -\frac{2}{2\times3} = -\frac{1}{3}$ - Vertex $y$ value: $y = 3(-\frac{1}{3})^2 + 2(-\frac{1}{3}) = 3(\frac{1}{9}) - \frac{2}{3} = \frac{1}{3} - \frac{2}{3} = -\frac{1}{3}$ - So vertex is $(-\frac{1}{3}, -\frac{1}{3})$ and AOS is $x = -\frac{1}{3}$ 5. For #4, $y = -9x^2 - 18x - 1$: - $a = -9$, $b = -18$ - $x = -\frac{-18}{2\times -9} = -\frac{-18}{-18} = -1$ - Vertex $y$: $-9(-1)^2 - 18(-1) - 1 = -9 + 18 - 1 = 8$ - Your vertex is $( -1, 26 )$, but calculation shows $y=8$, so this is incorrect. 6. For #5, $f(x) = -6x^2 + 24x - 20$: - $a = -6$, $b = 24$ - $x = -\frac{24}{2\times -6} = -\frac{24}{-12} = 2$ - Vertex $y$: $-6(2)^2 + 24(2) - 20 = -24 + 48 - 20 = 4$ - Vertex is $(2, 4)$ and AOS is $x=2$, which matches your answer. 7. For #6, $f(x) = 2x^2 + 8x + 4$: - $a=2$, $b=8$ - $x = -\frac{8}{2\times 2} = -2$ - Vertex $y$: $2(-2)^2 + 8(-2) + 4 = 8 - 16 + 4 = -4$ - Domain is all real numbers ($\mathbb{R}$) - Range is $y \geq -4$ since parabola opens upward - Your answers are correct. 8. For #7, $y = -8x^2 - 16x - 9$: - $a = -8$, $b = -16$ - $x = -\frac{-16}{2\times -8} = -\frac{-16}{-16} = -1$ - Vertex $y$: $-8(-1)^2 - 16(-1) - 9 = -8 + 16 - 9 = -1$ - Domain is $\mathbb{R}$ - Range is $y \leq -1$ since parabola opens downward - Your answers are correct. 9. For #8, $y = 3x^2 - 18x + 15$: - $a=3 > 0$ so parabola opens upward, vertex is minimum - Vertex $x = -\frac{-18}{2\times 3} = 3$ - Vertex $y = 3(3)^2 - 18(3) + 15 = 27 - 54 + 15 = -12$ - Max/Min: minimum, not maximum - Your max/min label is incorrect but vertex is correct. 10. For #9, $f(x) = -5x^2 + 10x + 7$: - $a = -5 < 0$ so parabola opens downward, vertex is maximum - Vertex $x = -\frac{10}{2\times -5} = 1$ - Vertex $y = -5(1)^2 + 10(1) + 7 = -5 + 10 + 7 = 12$ - Max/Min: maximum, not minimum - Your max/min label is incorrect but vertex is correct. 11. For #10, $y = 2x^2 - 10x + 13$: - $a=2 > 0$ so parabola opens upward, vertex is minimum - Vertex $x = -\frac{-10}{2\times 2} = \frac{10}{4} = \frac{5}{2}$ - Vertex $y = 2(\frac{5}{2})^2 - 10(\frac{5}{2}) + 13 = 2(\frac{25}{4}) - 25 + 13 = \frac{50}{4} - 12 = 12.5 - 12 = 1.5$ - Your vertex y-value is 1, but calculation shows 1.5 - Max/Min: minimum, not maximum 12. For #11, $h(t) = -16t^2 + 128t$: - Vertex time $t = -\frac{128}{2\times -16} = 4$ - Height at vertex: $h(4) = -16(4)^2 + 128(4) = -256 + 512 = 256$ - Your answers are correct. Summary: - Problems #1, #2, #3, #5, #6, #7, #11 are correct. - #4 vertex y-value is incorrect. - #8, #9 max/min labels are reversed. - #10 vertex y-value and max/min label are incorrect. Overall, you did well but check the vertex y-values and max/min labels carefully next time.